Question

The voltage and power ratings of a particular light bulb, which are its normal operating values, are 110 V and 60 W. Assume the resistance of the filament of the bulb is constant and is independent of operating conditions. If the light bulb is operated at a reduced voltage and the power drawn by the bulb is 36 W, what is the operating voltage of the bulb?

Answer 1

Answer:

The voltage operating is the 85,20563362

Explanation:

Power is the relation between Voltage and Current so knowing the resistance is going to be constant the equation of power can be just replacing in voltage terms:

1. $P=I*V$

using also Law OHM

2. $V= I*R$

$I = \frac{V}{R}$

Replacing 2 in the equation 1 so all the data are going to be in Voltage terms:

1. $P= \frac{V}{R} *V$

$P= \frac{V^{2} }{R}$  ⇒ $R= \frac{V^{2} }{P}$

$R= \frac{110^{2} v }{60 w}$

$R= 201,6666667$ Ω

So the resistance is constant so the current is going to be the same at the other Power 36 W:

$P= \frac{V^{2} }{R}$

$V^{2} = R*P$

$V=\sqrt{R*P}$

$V=\sqrt{201,6666667*36 }$

$V=\sqrt{7260}$ ⇒$V= 85,20563362$